Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b;
  cin >> a >> b;
  int M[a][b];
  int A[a];
  bool Ab[a];
  for (int i = 0; i < a; i++) {
    A[i] = Ab[i] = 0;
  }
  int B[b];
  bool Bb[b];
  for (int i = 0; i < b; i++) {
    B[i] = Bb[i] = 0;
  }
  int menor = 501;
  for (int i = 0; i < a; i++) {
    for (int j = 0; j < b; j++) {
      cin >> M[i][j];
      if (M[i][j] < menor) {
        menor = M[i][j];
      }
    }
  }
  vector<pair<int, bool> > V;
  for (int i = 0; i < a; i++) {
    for (int j = 0; j < b; j++) {
      M[i][j] -= menor;
      if (M[i][j] == 0) {
        if (Ab[i] == 0) {
          Ab[i] = 1;
          V.push_back(make_pair(i, true));
        }
        if (Bb[j] == 0) {
          Bb[j] = 1;
          V.push_back(make_pair(j, false));
        }
      }
    }
  }
  int ans = 0;
  vector<pair<int, int> > C;
  vector<pair<int, int> > R;
  int i = 0;
  while (i < V.size()) {
    if (V[i].second) {
      for (int j = 0; j < b; j++) {
        if (M[V[i].first][j]) {
          C.push_back(make_pair(j, M[V[i].first][j]));
          int x = M[V[i].first][j];
          ans += x;
          for (int k = 0; k < a; k++) {
            M[k][j] -= x;
            if (M[k][j] < 0) {
              cout << "-1" << endl;
              return 0;
            }
            if (M[k][j] == 0) {
              if (Ab[k] == 0) {
                Ab[k] = 1;
                V.push_back(make_pair(k, true));
              }
              if (Bb[j] == 0) {
                Bb[j] = 1;
                V.push_back(make_pair(j, false));
              }
            }
          }
        }
      }
    } else {
      for (int j = 0; j < a; j++) {
        if (M[j][V[i].first]) {
          R.push_back(make_pair(j, M[j][V[i].first]));
          int x = M[j][V[i].first];
          ans += x;
          for (int k = 0; k < b; k++) {
            M[j][k] -= x;
            if (M[j][k] < 0) {
              cout << "-1" << endl;
              return 0;
            }
            if (M[j][k] == 0) {
              if (Ab[j] == 0) {
                Ab[j] = 1;
                V.push_back(make_pair(j, true));
              }
              if (Bb[k] == 0) {
                Bb[k] = 1;
                V.push_back(make_pair(k, false));
              }
            }
          }
        }
      }
    }
    i++;
  }
  cout << ans + min(a, b) * menor << endl;
  for (int i = 0; i < min(a, b); i++) {
    for (int j = 0; j < menor; j++) {
      if (a < b) {
        cout << "row ";
      } else {
        cout << "col ";
      }
      cout << i + 1 << endl;
    }
  }
  for (int i = 0; i < C.size(); i++) {
    for (int j = 0; j < C[i].second; j++) {
      cout << "col " << C[i].first + 1 << endl;
    }
  }
  for (int i = 0; i < R.size(); i++) {
    for (int j = 0; j < R[i].second; j++) {
      cout << "row " << R[i].first + 1 << endl;
    }
  }
  return 0;
}